LeetCode 412. Fizz Buzz

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Description

Given an integer n, return a string array answer (1-indexed) where:

answer[i] == "FizzBuzz" if i is divisible by 3 and 5.
answer[i] == "Fizz" if i is divisible by 3.
answer[i] == "Buzz" if i is divisible by 5.
answer[i] == i (as a string) if none of the above conditions are true.

Example 1:

Input: n = 3
Output: ["1","2","Fizz"]
Example 2:

Input: n = 5
Output: ["1","2","Fizz","4","Buzz"]
Example 3:

Input: n = 15
Output: ["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]

Constraints:

1 <= n <= 104

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Analysis

Solved by enumeration.

Solution written in Rust

impl Solution {
    pub fn fizz_buzz(n: i32) -> Vec<String> {
        let mut res = Vec::new();
        for i in 1..=n {
            if i %3 == 0 && i % 5 == 0 {
                res.push("FizzBuzz".to_string());
            } else if (i % 3 == 0) { 
                res.push("Fizz".to_string());
            } else if (i % 5 == 0) {
                 res.push("Buzz".to_string());
            } else {
                res.push(i.to_string());
            }
        }
        res
    }
}

Solution written in Cpp

class Solution {
public:
    std::vector<std::string> fizzBuzz(int n) {
        std::vector<std::string> res;
        for (auto i = 1; i <= n; ++i) {
            if (i % 15 == 0) {
                res.emplace_back("FizzBuzz");
            } else if (i % 3 == 0) {
                res.emplace_back("Fizz");
            } else if (i % 5 == 0) {
                res.emplace_back("Buzz");
            } else {
                res.emplace_back(std::to_string(i));
            }
        }
        return res;
    }
};
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